can you check my work?
Solve the system of equations at the right by the method of substitution. Verify your results by using a graphing utility. x^2+y^2-4x+6y-5=0, x+y+5=0
x=-y-5
(-y-5)^2+y^2-4(-y-5)^2+6y-5=0
y=√5-5, y=-5-√5
x+√5-5+5=0
x=-√5
x-5-√5+5=0
x=√5
y=√5-5, y=-5-√5, x=-√5, x=√5
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Put the equations in, where they overlap is a solution to both.
To check the work, we first need to solve the system of equations using the method of substitution. The given equations are:
1) x^2 + y^2 - 4x + 6y - 5 = 0
2) x + y + 5 = 0
Let's solve equation 2) for x:
x = -y - 5
Now substitute this value of x into equation 1):
(-y - 5)^2 + y^2 - 4(-y - 5) + 6y - 5 = 0
Simplify the equation:
y^2 + 10y + 25 + y^2 + 4y + 20 + 6y - 5 = 0
2y^2 + 20y + 40 + 6y - 5 = 0
2y^2 + 26y + 35 = 0
We need to solve this quadratic equation. Factoring could be difficult, so let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = 26, and c = 35. Plugging in these values, we get:
y = (-26 ± √(26^2 - 4*2*35)) / (2*2)
y = (-26 ± √(676 - 280)) / 4
y = (-26 ± √396) / 4
y = (-26 ± 2√99) / 4
y = -6.5 ± 0.5√99
So the possible values for y are approximately -6.5 + 0.5√99 and -6.5 - 0.5√99.
Now we can substitute these values back into equation 2) to find the corresponding values of x.
For y = -6.5 + 0.5√99:
x = -(-6.5 + 0.5√99) - 5
x = 6.5 - 0.5√99 - 5
x = 1.5 - 0.5√99
For y = -6.5 - 0.5√99:
x = -(-6.5 - 0.5√99) - 5
x = 6.5 + 0.5√99 - 5
x = 1.5 + 0.5√99
So the solutions to the system of equations are:
(x, y) = (1.5 - 0.5√99, -6.5 + 0.5√99) and (x, y) = (1.5 + 0.5√99, -6.5 - 0.5√99).
Now, let's verify these results using a graphing utility. You can plot the two equations on a graph and see if they intersect at the same points we found.